Convergence of Series, Series Sums: sum[n=1,infty]cos(n pi)/

[Vempy]

I have two questions...

1) Sum of the series from n=1 to infinity of cos(n*pi)/2^n

I have that the series converges based on:

lim (n->inifinity) cos(n*pi)/2^n = cos(n*pi) * (1/2)^n = 0 because (1/2)^n -> 0 when n -> inifnity

Now I just don't know how to go about calculating the sum. The trig term is confusing me.


2) Determine whether or not sum of the series from n=1 to infinity of n^2/(n^4-6n^2+5)

I found that it might converge given:

lim (n->inifinity) n^2/(5n^2-3n-1) = (1/n^2)/(1-(6/n^2)+(5/n^4)) = 0/1 = 0

And then from there how do you make sure it converges? Do you use the integral test? If so I really don't know how to do that integral. X_x

 

[skeeter]

Re: Convergence of Series And Series Sums

I have two questions...

1) Sum of the series from n=1 to infinity of cos(n*pi)/2^n

I have that the series converges based on:

lim (n->inifinity) cos(n*pi)/2^n = cos(n*pi) * (1/2)^n = 0 because (1/2)^n -> 0 when n -> inifnity

Now I just don't know how to go about calculating the sum. The trig term is confusing me.

numerically list out a few terms of the series ... you should see a geometric series whose sum is easy to find.




2) Determine whether or not sum of the series from n=1 to infinity of n^2/(n^4-6n^2+5)

I found that it might converge given:

lim (n->inifinity) n^2/(5n^2-3n-1) = (1/n^2)/(1-(6/n^2)+(5/n^4)) = 0/1 = 0

And then from there how do you make sure it converges? Do you use the integral test? If so I really don't know how to do that integral. X_x

the nth term of a series approaching 0 as n goes to infinity does not guarantee convergence. consider the series 1/n ... the nth term goes to 0, but the series diverges.

I would use the limit comparison test with the known convergent series 1/n[sup:5jx4pdq5]2[/sup:5jx4pdq5] to establish convergence of your given series..

 

[Vempy]

Thank you.

So this is what I get:

1) The series, when written out, shows r = -1/2 therefore:

sum = -1/2 * 1/(1 - (-1/2))
= -1/3

2) Given a very large n the series looks like: 1/n^2, therefore:

S 1/n^2 = -n^-1 [1, infinity]
= 0 - 1
= -1

And because this series converges the other must as well.